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A heating element has resistance 100 $\Omega$ at 27.0 °C. Its temperature coefficient of resistance is $1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$. If the resistance rises to 117 $\Omega$, the temperature of the element is about
A640 °C
B847 °C
C1027 °C
D1227 °C
Answer & Solution
Correct answer: C. 1027 °C
1. Use $R_2 = R_1[1 + \alpha(T_2 - T_1)]$.
2. $T_2 - T_1 = (117 - 100)/(100 \times 1.70\times10^{-4}) = 17/0.017 = 1000\ ^\circ\text{C}$.
3. $T_2 = 1000 + 27 = 1027\ ^\circ\text{C}$.
4. 640 °C drops a factor; 1227 °C overshoots by adding too much to the rise.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.24_
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